Vtyjkyuk

Let $X$ be a random variable with values in $mathbbR$ and let $g: mathbbR to mathbbR$ a measurable function such that $ E g(X) $ exist.

If we define the expected value as $E(X)= int_x in mathbbRx cdot f_X(x) dx$, how can I deduce that $E(g(X))= int_x in mathbbRg(x) cdot f_X(x) dx$?

I know that this was asked before ( discrete case). Here Discrete case

My problem is that I cannot use the same technique here.

A fundamental theorem in measure theory is the key. Let $(Omega, F, P)$ be a probability space.
By definition we have $mathbbE g(X) = int_Omega g(X) dP$. We are aware that $g(X) = g circ X$. If $X: Omega to mathbbR$ is a random variable measurable-$F$ and if $p_X$ is the probability measure induced by $X$, then a fundamental theorem in measure theory (see, say the chapter on probability in Folland's real analysis) shows that
$$
int_Omega gcirc X dP = int_mathbbR g(x) dp_X(x).
$$
If $p_X$ is absolutely continuous with respect to the Lebesgue measure on $mathbbR$, then by the Radon-Nykodym theorem the probability density $f_X$ of $X$ exists, and a basic property of Lebesgue-Stieltjes integration asserts that
$$
int_mathbbR g(x)dp_X(x) = int_mathbbR g(x)f(x) dx.
$$